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JEE Main Online Exam 2019

Questions & Solutions 9th April 2019 | Shift - II

(Memory Based)

PHYSICS

Q.1 50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2 surface area will be close to (c = 3×108 m/s) :

(1) 20 × 10–8 N (2) 35 × 10–8 N (3) 10 × 10–8 N (4) 15 × 10–8 N Ans. [1]

Sol. F = ⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

cI

10075

cI2

10025

= ⎟⎠⎞

⎜⎝⎛

cI

100225

= 8103050

100125

××

= 20.8 × 10–8 N Q.2 The physical sizes of the transmitter and receiver antenna in a communication system are :

(1) inversely proportional to modulation frequency (2) proportional to carrier frequency (3) inversely proportional to carrier frequency (4) independent of both carrier and modulation frequency

Ans. [3] Sol. By theory size of atenna of reciever and transmitter both inverse to carrier frequency Q.3 A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion

constant 10–6 N-m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1° for a current of 1 mA. The value of B (in Tesla) is approximately :

(1) 10–3 (2) 10–1 (3) 10–4 (4) 10–2 Ans. [1]

Sol. τ = BM ×

C•θ = NIAB

10–6 180

π = 10–3 × 10–4 × 175 B

B = 10–3 T

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Q.4 Two coil 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P' a magnetic flux of 10–3 Wb passes through 'Q'. No current is passed through 'Q'. When no current passes through 'P' and a current of 2 A passes through 'Q', the flux through 'P' is :

(1) 6.67 × 10–3 Wb (2) 3.67 × 10–4 Wb (3) 6.67 × 10–4 Wb (4) 3.67 × 10–3 Wb Ans. [3] Sol. φ = BA

φP = 2/322

210

)xR(2Ri

+μ πr2 = 10–3

φQ = 2/322

220

)xr(2ri

+μ

πR2

2/3

22

22

1

2

q

P

xrxR

ii

⎟⎟⎠

⎞⎜⎜⎝

⎛

++

=φφ

φP = 3102 3−× = 6.67 × 10–4

Q.5 A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the

same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm', which move at angle 45° with respect to the original direction.

The speed of each of the moving particle will be :

(1) 2 ν (2) v/ )22( (3) ν/ 2 (4) 22 ν Ans. [4] Sol.

→2v m →v2m

before collision 45°m 45°

v=0

m

after collision m

Momentum conservation in x directon 2mv + 2mv = mv'cos 45° + mv' cos 45°

4v = 2 v'

v' = 2 2 v

Q.6 A wooden block floating in a bucket of water has 54 of its volume submerged. When certain amount of an oil

is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :

(1) 0.7 (2) 0.5 (3) 0.8 (4) 0.6 Ans. [4] Sol. For case 1 mg = F3

mg = m'g

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m = 54 Vdw

For case 2 mg = Fbw + Fbo

m = 2v dw +

2v do

By equation 1 & 2

54 Vdw =

2V dw +

2V do

106

dd

w

0 = = 0.6

Q.7 The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

(1) a + c4

b2 (2) a +

cb2

(3) a + c3

b2 (4) a +

c2b2

Ans. [3] Sol. x = at + bt2 – ct3 v = a + 2bt – 3 ct2 acceleration = 2b – 6ct = 0

f = c3b

c6b2

=

so velocity at t = c3b

v = a + 2b c3b – 3c 2

2

c9b

= a + c3

b2

Q.8 A thin smooth rod of length L and mass M is rotating freely with angular speed ω0

about an axis perpendicular to the rod and passing through its center. Two beads of mass m and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be :

(1) m3M

M 0

+ω (2)

m2MM 0

+ω (3)

mMM 0

+ω (4)

m6MM 0

+ω

Ans. [4] Sol.

J1=J2 (Angular momentum conservation)

12

M 2ω0 = ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

4m2

12M 22

ω

ω = m6M

M 0

+ω

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Q.9 A very long solenoid of readius R is carrying current I(t) = kte–αt (k > 0), as a function of time (t ≥ 0). Counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by :

(1)

I

t=0 t

(2)

I

t=0 t

(3)

I

t=0

t (4)

I

t=0 t

Ans. [2] Sol. φ = (μ0nKte–αt) 4πR2

e = dtdφ− = –ce–αt (1 – αt)

iinduced = R

)t1(ce t α−− α−

at t = 0 iind ⇒ +Vc Q.10 Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a

rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis for a duration of :

(1) 5 s (2) 3 s (3) 2.5 s (4) 2 s Ans. [4] Sol. ω1 = 0, α = 20 ω2 = ω1 + αt ω2 = 20 t

21 Iω2 = 1200

21 1.5 × 400 t2 = 1200

t = 2 sec Q.11 The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a

battery of voltage, V. When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is :

(1) nK

V+

(2) nK

nV+

(3) )nK(V)1n(

++ (4) V

Ans. [3] Sol. After full changing.

q2 = nC1V q1=C1V

q = CV + nCV = (n+1) CV Due to insertion of dielectric q1 = KCVC q2 = nCVC

VC = eff

total

Cq

= nK)1n(

nCKCCV)1n(

++

=+

+ V

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Q.12 The resistance of a galvanometer is 50 ohm and the maximum current which can be passed though it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0–0.5 A ?

(1) 0.2 ohm (2) 0.002 ohm (3) 0.02 ohm (4) 0.5 ohm Ans. [1]

Sol.

Ig G=50

S

Iy = 0.002 A S (0.5 – 0.002) = 50 × 0.002

S = 498.0

1.0 = 0.2

Q.13 The position vector of a particle changes with time according to the relation r (t) = 15t2 i + (4–20t2) j . What

is the magnitude of the acceleration at t = 1 ? (1) 100 (2) 40 (3) 50 (4) 25

Ans. [3] Sol. r (t) = 15t2 i + (4–20t2) j V = 30 t i – 40 tj acceleration = 30 i – 40 j

a = 22 )40(30 −+ = 50 Q.14 A massless spring (k = 800 N/m) attached with mass (500 g) is completely immersed in 1 kg of water. The

spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K)

(1) 10–3 K (2) 10–4 K (3) 10–5 K (4) 10–1 K Ans. [3]

Sol. 21 Kx2 = (m SdT)0 + (m SdT)w

21 800

2

1002

⎟⎠⎞

⎜⎝⎛ = (0.5×400+1×4184) dT

41042641600

×= dT

dT = 3 × 10–5 Order 10–5 K Q.15 Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness d and '3d' respectively,

are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are 'θ2' and 'θ1' respectivlely, (θ2 > θ1). The temperature at the interface is :

θ2 3K K

d 3d θ1

(1) 2

12 θ+θ (2) 3

23

21 θ+

θ (3) 109

1021 θ

+θ (4)

65

621 θ

+θ

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Ans. [3]

Sol. θ2 3K K

d 3d θ1

θ

Heat current will be same in both H1 = H2

dKA3 (θ2 – θ) =

d3KA (θ – θ1)

9θ2 – 9θ = θ – θ1

θ = 10

9 21 θ+θ

= 109

1021 θ

+θ

Q.16 Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a

speed of 20 ms–1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in car B?

(speed of sound in air = 340 ms–1) (1) 2060 Hz (2) 2150 Hz (3) 2300 Hz (4) 2250 Hz

Ans. [4]

Sol.

B A

source VA = 20 m/sec VB = 20 m/sec

Obs.

n' = n ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

s

0

vvvv

2000 = n ⎟⎠⎞

⎜⎝⎛

+−

2034020340

2000 = n ⎟⎠⎞

⎜⎝⎛

360320

n = 2000 × 89 = 2250 hr

Q.17 Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm. coming from a

distant object, the limit of resolution of the telescope if close to : (1) 2.0 × 10–7 rad (2) 4.5 × 10–7 rad (3) 1.5 × 10–7 rad (4) 3.0 × 10–7 rad

Ans. [4]

Sol. resolution limit+ = d22.1 λ

= 2

9

102501060022.1

−

−

×××

= 2.9 × 10–7 rad = 3.0 × 10–7 Q.18 The logic gate equivalent to the given logic circuit is :

A Y

B (1) AND (2) OR (3) NOR (4) NAND

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Ans. [2] Sol. A B 0 0 0 0 1 1 1 0 1 1 1 1 So it is OR gate

Q.19 A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index μ is put between the lens and the mirror, the pin has to be moved to A', such that OA' = 27 cm, to get

its inverted real image at A' itself. The value of μ will be :

MO

A

A'

L

(1) 3 (2)

34 (3)

23 (4) 2

Ans. [2]

Sol. 181

182

21

f1

1=×=

18

1f1

1 −−μ

=

P = 182

182

− (μ – 1)

Fm = – ⎟⎟⎠

⎞⎜⎜⎝

⎛μ−2

18

2 = 6 – 3 μ

μ = 34

Q.20 A wedge of mass M=4 m lies on a frictionless plane. A particle of mass m approaches the wedge with speed

ν. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by :

(1) g7

2 2ν (2) g2

2ν (3) g5

2 2ν (4) g

2ν

Ans. [3]

Sol.

m v 4 m

⇒ h v'

conservation of momentum mv = 5 mv'

v' = 5v

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energy conservation

21 mv2 =

21 5mv'2 + mgh

21 v2 =

25 2

5v

⎟⎠⎞

⎜⎝⎛ + gh

10v

2v 22

− = gh

h = g5v2 2

Q.21 The area of a square is 5.29 cm2. The area of 7 such squares taking into account the significant figures is :

(1) 37.03 cm2 (2) 37.0 cm2 (3) 37 cm2 (4) 37.030 cm2 Ans. [1] Sol. Total area = 7 × 5.29 = 37.03 cm2 special comment : Answer should be in two digit after decimal so answer should be (1), NTA give (4) Q.22 The specific heats, CP and CV of a gas of diatomic milecules, A, are given (in units of J mol–1 K–1) by 29 and

22, respectively. Another gas of diatomic moleucles, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then :

(1) Both A and B have a vibrational mode each (2) A is rigid but B has a vibrational mode. (3) A has a vibrational mode but B has none. (4) A has one vibrational mode and B has two.

Ans. [3] Sol. For A :

2229

CC

V

p =

1 + 2229

f2

=

227

f2

=

f = 744 = 6.6

3 translation, 2 rotation, remaining vibration For B :

1 + 2130

f2

=

219

f2

=

f = 942

No vibrational

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Q.23 In a conductor, if the number of conduction electrons per unit volume is 8.5 × 1028 m–3 and mean free time is 25 fs (femto second), it's approximate resistivity is :

(me = 9.1 × 10–31 kg) (1) 10–5 Ωm (2) 10–7 Ωm (3) 10–8 Ωm (4) 10–6 Ωm

Ans. [3] Sol. m = ne2 τρ

n = ρτ 2e

m

= 2821915

31

105.8)106.1(1025101.9

××××××

−−

−

= 1.67 × 10–8 Ω m Q.24 A He+ ion is in its first excited state. Its ionization energy is :

(1) 48.36 eV (2) 13.60 eV (3) 54.40 eV (4) 6.04 eV Ans. [2]

Sol. E = –13.6 2

2

nz eV

Z = 2 , n = 2 E = –13.6 eV ionisation energy is = +13.6 eV Q.25 A metal wire of resistance 3Ω is elongated to make a uniform wire of double its previous length. This new

wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be :

(1) 25

Ω (2) 35

Ω (3) 27

Ω (4) 5

12Ω

Ans. [2] Sol. R = 3 When length become double

R = 12 Ω , ⎟⎟⎠

⎞⎜⎜⎝

⎛ ρ=

ρ=

VAR

2

10 Ω

60° R ∝

REff = 610

= 35

Q.26 Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d, y = +2d, respectively.

The magnitude of the electric field E at a point a the x-axis at x = D, with D>>d, will behave as :

(1) D1E ∝ (2) 3D

1E ∝ (3) 4D1E ∝ (4) 2D

1E ∝

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Ans. [3] Sol. EP = 2E1 cos θ1 – 2E1 cos θ2

= 2/32222 )Dd(D

DdqK2

+×

+ 2/32222 ]D)d2[(

DD)d2(

qK2+

×+

−

= 2K qD [(d2+D2)–3/2 – (4d2 + D2)–3/2] d << D

= ⎟⎟⎠

⎞⎜⎜⎝

⎛ ×−−− 22

2

2 D2d431

Dd

231

DqDK2

= 4

2

DKqd9

Q.27 A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie

wavelengths 'λx' and 'λy' respectively. If x and y were moving in oppoiste directions, then the de-Broglie wavelength of 'P' is :

(1) λx – λy (2) yx

yx

λ−λ

λλ (3) λx + λy (4)

yx

yx

λ+λ

λλ

Ans. [2] Sol. Momentum conservation P1 + P2 = P

λ

=λ

−λ

hhh

21 ⇒ λ =

12

21

λ−λλλ

Q.28 A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third

harmonic mode. The speed of the wave and its fundamental frequency is : (1) 320 m/s, 80 Hz (2) 180 m/s, 120 Hz (3) 320 m/s, 120 Hz (4) 180 m/s, 80 Hz

Ans. [1]

Sol.

2 m

2V3 = 240

2V = 80 (fundumental frequency)

22

V×

= 80

V = 320 m/sec (velocity) Q.29 A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at

two distance x1 and x2 (x1 > x2) from the lens. The ratio of x1 and x2 is : (1) 4 : 3 (2) 3 : 1 (3) 2 : 1 (4) 5 : 3

Ans. [2] Sol. Magnification = 2

for x1 = 32f

x2 = 2f

2

1

xx = 3 : 1

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Q.30 A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρ(r) = 2rK .

Identify the correct relation between the radius R of the particle's orbit and its period T : (1) T/R2 is a constant (2) T/R is a constant (3) T2/R3 is a constant (4) TR is a constant

Ans. [2]

Sol.

M

2

2

rGMm

rmv

=

v = r

GM

dm = (4πr2dr) ρ

∫∫ π= 22

rK.drr4dm

m = 4π Kr

V = r

Kr4G π

V = KG4π

T = V

R2π

RT

→ const.

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CHEMISTRY

Q.1 The maximum number of possible oxidation states of actinoides are shown by : (1) nobelium (No) and lawrencium (Lr) (2) neptunium (Np) and lawrencium (Pu) (3) actinium (Ac) and thorium (Th) (4) berkelium (Bk) and californium (Cf)

Ans. [2] Sol.

77666655555

4444444433333333333333

LrNoMdFmEsCfBkCmAmPuNpUPaThAc

+++++++++++++++++

++++++++++++++

∴ Np and Pu has maximum number of possible O.S.

Q.2 In the following reaction carbonyl compound + MeOHHCl

acetal Rate of the reaction is the highest for : (1) Propanal as substrate and methanol in stoichiometric amount (2) Propanal as substrate and methanol in excess (3) Acetone as substrate and methanol in stoichiometric amound (4) Acetone as substrate and methanol in excess

Ans. [2]

Sol. Ketones < Aldehydes → Rate of NAR Only aldehydes are responsible for the function of acetals. Q.3 Among the following species, the diamagnetic molecule is :

(1) NO (2) O2 (3) CO (4) B2 Ans. [3] Sol. CO is diamagnetic in nature as all orbitals are fully filled.

CO MOT diagram = σ1s2 σ *1s2 σ2s2 σ*2s2 π2px2 = π2py

2 σ2pz2

∴ Ans. = 3

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Q.4 The correct statements among I to III regarding group 13 element oxides are, (I) Boron trioxide is acidic. (II) Oxides of aluminium and gallium are amphoteric. (III) Oxides of indium and thallium are basic.

(1) (I) and (II) only (2) (I) and (III) only (3) (II) and (III) only (4) (I), (II) and (III) only

Ans. [4]

Sol. B2O3 is acidic in nature Al2O3 and Ga2O3 are amphoteric, oxides of In and T are basic in nature. ∴ And. 4 Q.5 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many

mole of Ni will be deposited at the cathode? (1) 0.10 (2) 0.05 (3) 0.20 (4) 0.15

Ans. [2] Sol. gm E (Ni) = F Moles × V.F. = 0.1 x × 2 = 0.1 x = 0.05 Ni

+2 + 2e– → Ni Q.6 Molal depression constant for a solvent is 4.0 K kg mol–1. The depression in the freezing point of the solvent

for 0.03 mol kg–1 solution of K2OS4 is : (Assume complete dissociation of the electrolyte)

(1) 0.36 K (2) 0.18 K (3) 0.12 K (4) 0.24 K Ans. [1] Sol. ΔTf = ikfm = 3 (4) (0.03) ΔTf = 0.36 K Q.7 In an acid-base titration, 0.1 M HCl solution was added to the NaOH solution of unknown strength. Which of

the following correctly shows the change of pH of the titration mixture in this experiment?

pH

V(mL) (A)

pH

V(mL) (B)

pH

V(mL)(C)

pH

V(mL)(D)

(1) (C) (2) (B) (3) (A) (4) (D) Ans. [3] Sol.

HCl

NaOH → added

10 pH

3

NaOH

HCl → added

pH 10

3

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Q.8 The correct statements among I to III are : (I) Valence bond theory cannot explain the color exhibited by transition metal complexes. (II) Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes. (III) Valennce bond theory cannot distinguish ligands as weak and strong field ones. (1) (I), (II) and (III) (2) (II) and (III) only (3) (I) and (II) only (4) (I) and (III) only

Ans. [4] Sol. (1) VBT does not explain colour exhibited by complex of transition metal. (2) VBT does not distinguish between strong and weak complex. (3) VBT does not give quantitative interpretation of magnetic properties. ∴ According to question statement I and III are correct ∴ Ans. 4 Q.9 Increasing order of reactivity of the following compounds for SN1 substitution is :

CH2–Cl

CH3

CH3 (A)

Cl

(B)

H3C

(C) H3CO

Cl

(D)

Cl

(1) (A) < (B) < (D) < (C) (2) (B) < (C) < (D) < (A) (3) (B) < (C) < (A) < (D) (4) (B) < (A) < (D) < (C)

Ans. [4] Sol.

CH2–Cl

(C)

OCH3

>

CH2Cl (D)

> H3C–CH2–Cl (B)

> CH3–CH–CH2–Cl

(A)CH2

C > D > B > A Official answer is → 4 (for NTA) Fact → stability of carbocation ∝ Rate of SN2

Q.10 HF has highest boiling point among hydrogen halides, because it has :

(1) lowest ionic character (2) lowest dissociation enthalpy (3) strongest van der Waals intreactions (4) strongest hydrogen bonding

Ans. [4] Sol. HF has strong hydrogen bond ∴ it has highest B.P. among hydrogen halides. ∴ Ans. 4

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Q.11 The major product of the following reaction if :

CH2OH

OH

CO2Et

H2SO4(cat.)CHCl3

(1)

COOH

O

(2)

CO2Et

O

(3)

OH

O

OEt (4)

OH

O

O

Ans. [4] Sol.

OH OH

C–O–CH2–CH3

O

H2SO4

Hydrolysis of ester

OHOH

O

C–O–H

+H⊕

–H2O

OH

O

O

Q.12 The maximum possible denticities of a ligand given below towards a common transition and inner-transition

metal ion, respectively, are :

N ΘOOC

ΘOOC N N

COOΘ

COOΘCOOΘ

(1) 6 and 6 (2) 8 and 8 (3) 8 and 6 (4) 6 and 8 Ans. [4] Sol. General C.No. of CN– in transition elements is 6 and in inner transition elements is 8 – 12. ∴ ans. 4 Q.13 Noradrenaline is a / an :

(1) Antacid (2) Antihistamine (3) Antidepressant (4) Neurotransmitter Ans. [4] Sol. Nonadrenaline is a neurotransmitter.

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Q.14 What would be the molality of 20 % (mass/mass) aqueous solution of KI ? (molar mass of KI = 166 g mol–1)

(1) 1.51 (2) 1.35 (3) 1.08 (4) 1.48 Ans. [1] Sol. 20% w/w K.I. 20 gm K.I. is 100 gm solution

Molality = )sovlent(kgmw

)solute(gm×

= 80166

100020×

×

= 1.506 = 1.51 Q.15 The peptide that gives positive ceric ammonium nitrate and carbylamine tests is :

(1) Gln - Asp (2) Asp - Gln (3) Ser - Lys (4) Lys - Asp Ans. [3] Sol. Ser-lys Due to –OH group of serine it gives ⊕ve serric ammonium nitrates tent whereas due to –NH2 group lysine

gives ⊕ve. Carbylamine test. Q.16 10 mL of 1 mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar

head is approximated as a cube, what is its edge length? (1) 2.0 nm (2) 0.1 nm (3) 2.0 pm (4) 1.0 pm

Ans. [3] Sol.

Moles = 1000mvml =

10001010 3 ×−

= 10–5 mole

10–5 NA molecules covering area = 0.24 cm2

1 NA molecules covering area = A

5N1024.0

−cm2

235 1061024.0

××− = a2

a2 = 4 × 10–20 cm2 a = 2 × 10–10 cm a = 2 × 10–12 m a = 2 pm

Q.17 Which of the following compounds is a constituent of the polymer

– HN–C–NH–CH2 –

O

N ?

(1) N-Methyl urea (2) Ammonia (3) Formaldehyde (4) Methylamine Ans. [3] Sol.

Urea Formaldehyde H2N–C–NH2+H–C–H

O

[ NH–C–NH–CH2 ]n

O Urea formaldehyde resin

O

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Q.18 p-Hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives :

(1)

Br O

HO (2)

Br O

HO

Br

(3)

Br

O

HO

(4)

Br

O

HO

Ans. [3] Sol.

O

HO

Br/CCl4

(M effect)

OBr

HO

+M effect having group is responsible for activation towards ESR. Q.19 Hinsberg' reagent is :

(1) SOCl2 (2) (COCl)2 (3) C6H5COCl (4) C6H5SO2Cl Ans. [4] Sol. Benzene sulphonyl chloride C6H5SO2C

O

S

O

Cl

Q.20 Which of the following potential energy (PE) diagrams represents the SN1 reaction ?

(1)

Progress of reaction →

↑ PE (2)


↑ PE

(3)


↑ PE

(4)


↑ PE

Ans. [2] Sol. For SN1 carbocation is formed.

Rate of reaction →

P.S.↑ Inter mediate

F.S. T.S

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Q.21 The structures of beryllium chloride in the solid state and vapour phase, respectively, are : (1) chain and chain (2) dimeric and dimeric (3) dimeric and chain (4) chain and dimeric

Ans. [4] Sol. Solid state-chain

–Be

Cl

Cl Be

Cl

Cl –Be

In vapour state dimeric

Cl–Be

Cl

Cl Be–Cl

Q.22 Assertion : For the extraction of iron, haematite ore is used. Reason : Haematite is a carbonate ore of iron.

(1) Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion. (2) Both the assertion and reason are correct and the reason is the correct explanation for the assertion. (3) Only the reason is correct. (4) Only the assertion is correct.

Ans. [4] Sol. Extration of Fe is done from Haematite ore this is true but reason is wrong as Haematite is Fe2O3. ∴ Ans. 4 Q.23 During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The

change in internal energy, ΔU (in kJ) is : (1) 12 (2) –8 (3) 8 (4) –12

Ans. [3] Sol. Work done on system = +10 kT Heat escaped = –2kI ΔU = q + w = 10 – 2 = 8 KJ. Q.24 At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their

equation of state is given as P = bV

RT−

at T.

(1) Kr (2) Ar (3) Xe (4) Ne Ans. [3]

Sol. P = )bV(

RT−

P(V–b) = RT

⎟⎠

⎞⎜⎝

⎛ + 2VaP (V – b) = RT

At high pressure. P(V – b) = RT PV – Pb = RT

RTPV –

RTPb = 1

Z = 1 + RTPb

Z > 1, Z ∝ b

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Q.25 The one that is not a carbonate ore is : (1) malachite (2) bauxite (3) calamine (4) siderite

Ans. [2] Sol. Malachite = CuCO3. Cu(OH)2 Bauxite = Al2O3 . 2H2O or AlOx . (OH)3–2x (where 0 < x < 1) Calamine = ZnCO3 Siderite = FeCO3 ∴ Ans is Bauxite option-2. Q.26 The amorphous from of silica is :

(1) kieselguhr (2) tridymite (3) cristobalite (4) quartz Ans. [1] Sol. Kieselguhr is amorphous form of silica. ∴ Ans. 1 Q.27 Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect ? (The

Bohr radius is represented by a0). (1) The magnitude of the potential energy is double that of its kinetic energy on an average. (2) The probability density of finding the electron is maximum at the nucleus. (3) The total energy of the electron is maximum when it is at a distance a0 from the nucleus. (4) The electron can be found at a distance 2a0 from the nucleus.

Ans. [3]

Sol. (1) T.E. = -K.E. = 2PE

(4)

Is does not have any boundary. Q.28 The layer of atmosphere between 10 km to 50 km above the sea level is called as :

(1) troposphere (2) thermosphere (3) stratosphere (4) mesosphere Ans. [3] Sol. Memory based. Q.29 Consider the given plot of enthalpy of the following reaction between A and B. A + B → C + D. Identify the incorrect statement.

2015105 A+B C

D

Reaction Coordinate

Enthalpy(kJ mol–1)

(1) C is the thermodynamically stable product. (2) Formation of A and B from C has highest enthalpy of activation. (3) Activation enthalpy to form C is 5 kJ mol–1 less than that to form D. (4) D is kinetically stable product.

Ans. [3] Sol. Ea = (d → c) = 15 – 0 = 15 Ea = (A + B) → C = 15 Ea = (A + B → D) = 10 Ea = C → (A + B) = 20

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Q.30 The major products A and B for the following reactions are, respectively :

[A] KCN DMSO

H2/Pd [B]

O I

(1)

;

CN HO CH2–NH2HO HI

(2)

;

CH2NH2CN O O

(3)

;

CH2NH2CN O OH

(4)

;

CN HO CH2–NH2HO II

Ans. [3] Sol.

O x

KCN/DMSO SN2

O CN

H2/Pd

OH CH2–NH2

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(Memory Based)

MATHEMATICS

Q.1 If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and 2x – y + z = 0 has a non-trivial solution (x, y, z),

then xz

zy

yx

++ + k is equal to :

(1) 21 (2)

43 (3)

41

− (4) –4

Ans. [1] Sol. Given system of equations has non-trival solution

∴ Δ = 0 ⇒ 1122K1132

−−−

= 0

⇒ K = 29

so equation are 2x + 3y – z = 0 .........(1)

x + 29 y – 2z = 0 .........(2)

2x – y + z = 0 .........(3) (1) – (3) ⇒ 4y – 2z = 0 ⇒ 2y = z .........(4)

⇒ 21

zy

=

From equation (1) & (4) 2x + 3y – 2y = 0 ⇒ 2x + y = 0

⇒ 21

yx −

=

or 4xz

−=

21K

xz

zy

yx

=+++

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Q.2 The common tangent to the circles x2 + y2

= 4 and x2 + y2

+ 6x + 8y – 24 = 0 also passes through the point : (1) (–6, 4) (2) (4, –2) (3) (–4, 6) (4) (6, –2)

Ans. [4] Sol. circle x2 + y2 = 4 ⇒ c1(0, 0) ; r1 = 2 and circle x2 + y2 + 6x + 8y – 24 = 0 ⇒ c2 (–3, –4) ; r2 = 7 ⇒ d = c1c2 = 5 also d = |r1 – r2| circles touch externally equation of common tangent s1 – s2 = 0 ⇒ 6x + 8y – 20 = 0 ⇒ 3x + 4y – 10 = 0 Point (6, –2) satisfy it Q.3 Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball,

the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is :

(1) 157 (2) 262 (3) 225 (4) 190 Ans. [4]

Sol. 2

)1n(n + + 99 = (n – 2)2

⇒ n2 + n + 198 = 2n2 – 8n + 8 ⇒ n2 – 9n – 190 = 0 ⇒ (n – 19) (n + 10) = 0 ⇒ n = 19

number of balls = 220.19 = 190

Q.4 If ( )∫ ++ dx)xsecxtanx(sec)x(fxtanxsece 2xsec = esecx f(x) + C, then a possible choice of f(x) is :

(1) secx – tanx – 21 (2) secx + xtanx –

21

(3) secx + tanx + 21 (4) xsecx + tanx +

21

Ans. [3]

Sol. ∫ xsece (secx tanx f(x) + (secx tanx + sec2x)) dx

= esecx f(x) + C Diff both side w.r.t x ⇒ esecx (secx tanx f(x) + (secx tanx + sec2x)) = esecx . secx tanx f(x) + esecx f '(x) ⇒ f '(x) = sec2x + tanx secx ⇒ f(x) = tanx + sec + C

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Q.5 If the tangent to the parabola y2 = x at a point (α, β), (β > 0) is also a tangent to the ellipse, x2 + 2y2 = 1, then α is equal to :

(1) 2 + 1 (2) 2 2 + 1 (3) 2 – 1 (4) 2 2 – 1 Ans. [1] Sol. Equation of tangent to the parabola y2 = x at (α, β) is T = 0

yβ = 2

x α+

⇒ yβ = 2

x 2β+ (∴ β2 = α)

⇒ y = 2

x21 β

+β

⎟⎟⎠

⎞⎜⎜⎝

⎛ β=

β=

2c,

22m

this is also a tangent to ellipre x2 + 2y2 = 1

∴ C = 222 bma +±

⇒ 21

41

2 2 +β

±=β

⇒ 21

41

4 2

2+

β=

β

⇒ β 4 – 2β2 – 1 = 0 ⇒ (β – 1)2 = 2 ⇒ β 2 – 1 = 2 (∵ β > 0)

⇒ β 2 = 2 + 1 α = β2 = 2 + 1

Q.6 The total number of matrices A = ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−1yx21yx2

1y20, (x, y∈R, x ≠ y) for which ATA = 3I3 is :

(1) 2 (2) 4 (3) 3 (4) 6 Ans. [2] Sol. (AT) (A) = 3I3

⇒ ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

300030003

1yx21yx2

1y20

111yyy2x2x20

⇒ ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

300030003

3000y6000x8

2

2

⇒ 8x2 = 3 ⇒ x = ± 83

⇒ 6y2 = 3 ⇒ y = ±21

4 materices are possible

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Q.7 If f : R → R is a differentiable function and f(2) = 6, then ∫ −→

)x(f

62x )2x(

dtt2lim is :

(1) 0 (2) 24 f ' (2) (3) 12 f ' (2) (4) 2 f ' (2) Ans. [3]

Sol. ∫ −→

)x(f

62x

dx)2x(

tdt2lim {given that f(2) = 6}

00 form. So we use L-Hospital Rule

= 1

)x(f2).x('flim2x→

= f '(2) . 2f(2) = 12f ' (2) Q.8 If a unit vector a makes angles π/3 with i , π/4 with j and θ ∈(0, π) with k , then a value of θ is :

(1) 125π (2)

65π (3)

4π (4)

32π

Ans. [4] Sol. cos2 α + cos2 β + cos2 γ = 1

⇒ 21

41

+ + cos2 γ = 1

⇒ cos2γ = 1 – 41

43

=

⇒ cosγ = ±21

⇒ γ = 3

2or3

ππ

Q.9 The area (in sq. units) of the smaller of the two circles that touch the parabola, y2 = 4x at the point (1, 2) and

the x-axis is : (1) 4π (2 – 2 ) (2) 8π (3 – 2 2 ) (3) 4π (3 + 2 ) (4) 8π (2 – 2 )

Ans. [2] Sol.

C(3–r, r)

r

(3 – r, 0)

x

x+y–3=0

0

yy=x+1

P(1, 2)

equation of tangent to the perabola y2 = 4x at (1, 2) is

2y = 4 ⎟⎠⎞

⎜⎝⎛ +

21x

⇒ y = x + 1

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equation of normal y = –x + 3 Let center be c(3 – r, r) Now PC2 = r2 ⇒ (3 – r – 1)2 + (r – 2)2 = r2 ⇒ 2(2 – r)2 = r2

⇒ r2 – 8r + 8 = 0 ⇒ r = 4± 2 2 for r = 4 + 2 2 (3 – r < 0) Not possible So r = 4 – 2 2 Area = πr2 = π(16 + 8 – 16 2 ) = 8π (3 – 2 2 ) Q.10 Two newspapers A and B are published in a city. It is known that 25 % of the city population reads A and

20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisements is :

(1) 13.9 (2) 13.5 (3) 12.8 (4) 13 Ans. [1] Sol. Let population = 100 n(A) = 25 n(B) = 20 n(A ∩ B) = 8 n(A ∩ B ) = 17 n( A ∩ B) = 12

A B

17 8 12

Now % of the population who look advertiesement

= 81005012

1004017

10030

×+×+×

= 13.9 Q.11 The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + .... upto 11th term is :

(1) 946 (2) 916 (3) 915 (4) 945 Ans. [1] Sol. S = 1 + 2 × 3 + 3 × 5 + 4 × 7 + ......+ upto 11th terms nth term of the sevies is Tn = n(2n – 1)

⇒ S = ∑∑==

−=11

1n

211

1n

)nn2(Tn

⇒ Sn = 2

)1n(n6

)1n2)(1n(n2 +−

++

put n = 1

⇒ S11 = 2

)12(116

)23)(12)(11(2−

⇒ S11 = 946

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Q.12 If the function f(x) = ⎩⎨⎧

>+π−≤+−π

5x,3xb5x,1xa

is continous at x = 5, then value of a – b

(1) 5

2+π

(2) 5

2+π

− (3) 5

2−π

(4) π−5

2

Ans. [4]

Sol. f(x) = ⎩⎨⎧

>+−π<+−π

5x,3|x|b5x,1|x|a

cortinous at x = 5 ∴ L.H.L = R.H.L = f(5) ⇒ b|π – 5| + 3 = a|π – 5| + 1 ⇒ –b(π – 5) +3 = –a(5 – π) + 1 ⇒ (a – b) (π – 5) = –2

⇒ a – b = π−

=−π

−5

25

2

Q.13 Let z ∈ C be such that |z| < 1. If ω = )z1(5

z35−

+ then :

(1) 4 Im (ω) > 5 (2) 5 Re (ω) > 1 (3) 5 Im (ω) < 1 (4) 5 Re (ω) > 4 Ans. [2]

Sol. ω = )z1(5

325−

+

⇒ 5ω – 5ωz = 5 + 3z

⇒ z = ω+

−ω53

55

given |z| < 1

⇒ ω+

−ω53

55 < 1

⇒ |5ω – 5 | < |3 + 5ω|

⇒ |ω – 1| < ω+43

53

−

Re(ω) = 51

∴Re(ω) > 51

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Q.14 If f(x) = [x] – ⎥⎦⎤

⎢⎣⎡

4x , x ∈ R, where [x] denotes the greatest integer function, then :

(1) −→4x

lim f (x) exists but +→4x

lim f (x) does not exist.

(2) Both −→4x

lim f (x) and +→4x

lim f (x) exist but are not equal.

(3) +→4x

lim f (x) exists but −→4x

lim f (x) does not exist.

(4) f is continuous at x = 4. Ans. [4]

Sol. f(x) = [x] – ⎥⎦⎤

⎢⎣⎡

4x

+→4x

lim f(x) = +→4x

lim⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−

4x]x[ = 4 – 1 = 3

−→4x

lim f(x) = −→4x

lim ⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡−

4x]x[ = 3 – 0 = 3

f(4) = 3 ∴ continous at x = 4 Q.15 A rectangle is inscribed in a circle with a diameter lying along the line 3y = x + 7. If the two adjacent vertices

of the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq. units) is : (1) 98 (2) 56 (3) 72 (4) 84

Ans. [4]

Sol. C D

A B (–8, 5) (6, 5)

(6, K) 3y = x + 7

Let vertex c is (6, K) then center of circle ⎟⎠⎞

⎜⎝⎛ +

−2

K5,1 it lies on diameter 3y = x + 7

⇒ 3 ⎟⎠⎞

⎜⎝⎛ +

2K5 = –1 + 7

⇒ K = –1 So AB = 14 and BC = 6 Area = 14 × 6 = 84

Q.16 The domain of the definition of the function f(x) = 2x41

−+ log10 (x3 – x) is :

(1) (1, 2) ∪ (2, ∞) (2) (–2, –1) ∪ (–1, 0) ∪ (2, ∞) (3) (–1, 0) ∪ (1, 2) ∪ (2, ∞) (4) (–1, 0) ∪ (1, 2) ∪ (3, ∞)

Ans. [3]

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Sol. f(x) = 2x41

−+ log10

(x3 – x)

Let f1 = 2x41

− and f2 = log10 (x3 – x)

⇒ 4 – x2 ≠ 0 x3 – x > 0 ⇒ x ≠ ±2 ⇒ x(x – 1) (x + 1) > 0

⇒ – + ––1 0

+1

x∈(–1, 0 ) ∪ (1, ∞) – {2} x∈(–1, 0 ) ∪ (1, 2) ∪ (2, ∞) Q.17 Two poles standing on a horizontal ground are of heights 5 m and 10 m respectively. The line joining their

tops makes and angle of 15° with the ground. Then the distance (in m) between the poles, is :

(1) 10 ( 3 –1) (2) 5(2 + 3 ) (3) 5( 3 + 1) (4) 25 (2+ 3 )

Ans. [2]

Sol.

D E

A

B

x

5 cm 10 cm 15°

5 C

in ΔABC ⇒ tan15° = x5

⇒ 2 – 3 = x5

⇒ x = 5(2 + 3 ) Q.18 If m is chosen in the quadratic equation (m2

+ 1) x2 – 3x – (m2 + 1)2 = 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :

(1) 38 (2) 34 (3) 510 (4) 58

Ans. [4] Sol. (m2 + 1)x2 – 3x + (m + 1)2 = 0

⇒ α + β = 1m

32 +

αβ = 1m)1m(

2

2

++

∵ α + β is maximum ∴ m2 + 1 is minimum ⇒ m = 0 ∴ α + β = 3 and αβ = 1 |α3 – β3| = |(α – β) (α2 + αβ + β2)|

= { }αβ−β+ααβ−β+α 22 )(4)(

= )19(49 −−

= 58

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Q.19 The value of the integral ∫ −1

0

1cotx (1–x2+x4) dx is :

(1) 21

4−

π loge 2 (2) 2π – loge 2 (3)

4π – loge 2 (4)

21

2−

π loge 2

Ans. [1]

Sol. I = ∫ −1

0

1cotx (1 – x2 + x4) dx

I = ⎟⎠⎞

⎜⎝⎛

+−∫ −42

1

0

1

xx11tanx dx

I = ⎭⎬⎫

⎩⎨⎧

−+−−

∫ −

)1x(x1)1x(xtanx 22

221

0

1 dx

I = ∫1

0

x {tan–1 x2 – tan–1 (x2 – 1)}dx

let x2 = t ⇒ 2x dx = dt

I = { }∫ −− −−1

0

11 )1t(tanttan21 dt

= ∫∫ −− −−1

0

11

0

1 )1t(tan21dtttan

21 dt

= ∫∫ −− −−1

0

11

0

1 )t(tan21tdttan

21 dt

= ∫∫ −− +1

0

11

0

1 )t(tan21dtttan

21 dt

= ∫ −1

0

1 )t(tan dt

= ∫ +−−

1

02

10

1

t1t)tan.(tan dt

= [ ]102t1log(21

4+−⎟

⎠⎞

⎜⎝⎛ π

= 2log21

4 e−π

Q.20 Let P be the plane, which contains the line of intersection of the planes, x + y + z – 6 = 0 and 2x + 3y + z + 5

= 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to : (1) 205 5 (2) 11/ 5 (3) 63 5 (4) 17/ 5

Ans. [2] Sol. Equation of plane P1 + λ P2 = 0 (x + y + z – 6) + λ(2x + 3y + z + 5) = 0 ⇒ x(1 + 2λ) + y(1 + 3λ) +z (1 + λ) – 6 + 5λ = 0 This plane is ⊥ to xy - plane ∴ 1 + λ = 0 ⇒ λ = –1

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So, equation of plane –x – 2y – 11 = 0 ⇒ x + 2y + 11 = 0 distance of the point (0, 0, 256) from this plane

= 5

1141100

=+

++

Q.21 If some three coefficients in the binomial expansion of (x + 1)n in powers of x are in the ratio 2 : 15 : 70,

then the average of these three coefficients is : (1) 232 (2) 964 (3) 625 (4) 227

Ans. [1]

Sol. given : 152

1rnr

152

CC

rn

1rn

=+−

⇒=−

⇒ 15r = 2n – 2r + 2 ⇒ 17r = 2n + 2 ......(1)

also given 143

rn1r

7015

CC

1rn

rn

=−+

⇒=+

⇒ 3n – 3r = 14r + 14 ⇒ 17r = 3n – 14 Solving (1) & (2) n = 16, r = 2

average of coefficient = 3

CCC 316

216

116 ++

= 3

56012016 ++

= 232

Q.22 A water tank has the shape of an inverted right cirular cone, whose semi-vertical angle is tan–1 ⎟⎠⎞

⎜⎝⎛

21 . Water is

poured into it at a constant rate of 5 cubic meter per minute. Then the rate (in m/min.), at which the level of water is rising at the instant when the depth of water in the tank is 10 m; is :

(1) 1/5 π (2) 1/10 π (3) 1/15 π (4) 2/π Ans. [1]

Sol. given θ = tan–1 ⎟⎠⎞

⎜⎝⎛

21

r

θ h = 10m

⇒ tanθ =

hr

21

=

⇒ r = 2h

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v = 31

πr2h

v = 31

π 4h3

dtdh)h3(

12dtdv 2π

=

S = π

=⇒π

51

dtdh

dtdh)100(

4

Q.23 The value of sin10° sin30° sin50° sin70° is :

(1) 161 (2)

361 (3)

181 (4)

321

Ans. [1] Sol. = sin10° sin30° sin50° sin70° = sin30°{sin10° sin(60°–10°) sin(60° + 10°)}

= sin30° ⎭⎬⎫

⎩⎨⎧ °)10(3sin

41

= ⎟⎠⎞

⎜⎝⎛ ×

21

41

21

= 161

Q.24 If p ⇒ (q ∨ r) is false, then the truth values of p, q, r are respectively : (1) F, F, F (2) T, F, F (3) F, T, T (4) T, T, F

Ans. [2] Sol. p ⇒ (q ∨ r) is false. (∵ T ⇒ F = F) So, P = T, q = F and r = F

Q.25 The vertices B and C of ΔABC lie on the line, 4z

01y

32x

=−

=+ such that BC = 5 units. Then the area (in sq.

units) of this triangle, given that the point A(1, –1, 2), is :

(1) 5 17 (2) 34 (3) 2 34 (4) 6

Ans. [2]

Sol.

A(1, –1, 2)

D 5

B λ==

−=

+4z

01y

32x

C

Let any point on given line is D(3λ –2, 1, 4λ) Now AD ⊥ BC D.R. of BC ⇒ a1 = 3, b1 = 0, c1 = 4 D.R. of AD ⇒ a2 = 3λ – 3, b2 = 2, c2 = 4λ – 2

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⇒ a1a2 + b1b2 + c1c2 = 0 ⇒ 3(3λ – 3) + 0 + 4(4λ – 2) = 0 ⇒ 25λ = 17

⇒ λ = 2517

co-oridnation of point D ⎟⎠⎞

⎜⎝⎛

2568,1,

251

AD = 3452

253244

625576

=++

Area of ΔABC = 21 × BC × AD

= 21 × 5 × 34

52

= 34

Q.26 If cosx dxdy – ysinx = 6x, (0<x<

2π ) and y ⎟

⎠⎞

⎜⎝⎛ π

3= 0, then y ⎟

⎠⎞

⎜⎝⎛ π

6is equal to :

(1) – 34

2π (2) – 2

2π (3) 32

2π (4) – 32

2π

Ans. [4]

Sol. cosx dxdy – ysinx = 6x

⇒ dxdy – y tanx = 6x sec x

I.F = xsec

1ee xseclogdxxtane == −−∫

∴ solution of equation

⇒ y. ∫= dxxsec

1.xsecx6xsec

1

⇒ xsec

y = 3x2 + c ........(1)

given y ⎟⎠⎞

⎜⎝⎛ π

3= 0

So, 0 = 9

3 2π + C

⇒ C = – 3

2π

Now from (1)

⇒ 3

x3xsec

y 22 π

−=

at x = 6π

⇒ 336

32

y3 22 π−

π=

⇒ y = – 32

2π

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Q.27 If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th term is :

(1) –25 (2) –35 (3) –36 (4) 25 Ans. [1] Sol. Let the three numbers in A.P. are a – d, a, a + d given that : a – d + a + d = 33 ⇒ a = 11 and (a – d)(a)(a + d) = 1155 ⇒ a(a2 – d2) = 1155 ⇒ 11(121 – d2) = 1155 ⇒ d2 = 16 ⇒ d = ±4 If d = 4 then first term a – d = 7 If d = –4 then first term a – d = 15 T11 = 7 + 10(4) = 47 or T11 = 15 + 10(–4) = –25 Q.28 The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34 x, 42, 67, 70, y

are 42 and 35 respectively, then xy is equal to :

(1) 7/2 (2) 9/4 (3) 8/3 (4) 7/3 Ans. [4] Sol. mean = 42

⇒ 10

y706742x3429262210 +++++++++ = 45

⇒ 420 = 300 + x+ y ⇒ x + y = 120 ......(1) and medium = 35

⇒ 2

x34 + = 35

⇒ x = 36 From (1) y = 84

37

3684

xy

==

Q.29 If the two lines x + (a – 1) y = 1 and 2x + a2y = 1 (a∈ R –{0, 1}) are perpendicular, then the distance of their

point of intersection from the origin is :

(1) 52 (2)

52 (3)

52 (4)

52

Ans. [2] Sol. Two lines are perpendicular ∴ m1m2 = –1

⇒ ⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛

−−

2a2

1a1 = –1

⇒ a3 – a2 + 2 = 0 ⇒ (a + 1) (a2 – 2a + 2) = 0

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∴ a = –1 so lines are L1 : x – 2y + 1 = 0 L2 : 2x + y – 1 = 0

Solving there equation we get point of intersetion P ⎟⎠⎞

⎜⎝⎛

53,

51

Now distance of P from origin

OP = 52

259

251

=+

Q.30 The area (in sq. units) of the region A = {(x, y) : 2y2

≤ x ≤ y + 4} is :

(1) 18 (2) 30 (3) 3

53 (4) 16

Ans. [1] Sol.

(8, 4)

0

y

(2, –2)

x

y2 = 2x .......(1) and x – y – 4 = 0 .......(2) solving (1) & (2) (x – 4)2 = 2x ⇒ x2 – 10x + 16 = 0 ⇒ x = 81 + 2 and y = 41 –2

A = ∫−

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

4

2

2

2y4y dy

A = 4

2

32

3yy4

4y

−⎟⎟⎠

⎞⎜⎜⎝

⎛−+

A = ⎟⎠⎞

⎜⎝⎛ +−−⎟

⎠⎞

⎜⎝⎛ −+

6881

664164 = 18

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